**Linear equation in one variable **

*Introduction: **Variable : A symbol which can takes various numerical values is called variable. *

*Ex :* x,y…

*Term : A Constant alone or A variable alone or combination of constant and variable with multiplication* ‘×’ ,* division* ‘÷’* is called term.*

*Multiplication* ‘×’ ,*division* ‘÷’* do not separate the terms.*

*Ex: x/2y, *xy *are single terms.*

*Expression : An expression is a single term or a combination of terms connected by the symbols* ‘+’*(plus)* *or* ‘-‘* (minus)*.

*Ex* : x²+y *is an expression having two terms.*

x+y²+z *is an expression having 3 terms.*

*Algebraic Equation : An algebraic equation is equality of algebraic expression involving variables and constants. It has equality sign. Left side of the equality sign is called LHS. Right side of equality is called RHS.*

*In an equation the values of LHS and RHS are equal. This happens to be true only for certain value of the variable. This is called the solution of the equation.*

x+7=8 *is an equation. Solution is* x=1.

*If degree of equation is* ‘1’ *then it is called linear equation.*

*Ex*: x+5=6

x²+4x+5=0 *is not a linear equation because its degree is* 2

x⁴+7x=9* is not a linear equation because it’s degree is* 4.

*Transposition : In a equation any constant ‘or’ variable ‘or’ term may be taken to the other side with a change in its sign ‘or’ position. This process is called transposition. *

*How to solve an equation:*

1)*We can add the same number on both sides of the equation.*

2)*We can subtract the same number from both sides of the equation.*

3)*We can multiply both sides of the equation with same number.*

4)*We can divide both sides of the equation by the same non-zero number.*

*When we transpose terms:*

*‘+’ term becomes ‘ – ‘ term.*

*‘ – ‘ term becomes’+’term.*

*Multiplied quantity becomes divided quantity.*

*Divided quantity becomes multiplied quantity.*

*P*) x-2=7* then find* ‘x’ *value.*

*Ans: From what number* ‘2’* should be subtract to get* ‘7’.

*i.e.* x=9 * ‘or’ *

*Add* 2 *on both sides. Then we get* x=9.

*P) Solve* 7x-9=16.

*solution: Given* 7x-9=16.* *

*Add* ‘9’ * on both sides.*7x-9+9=16+9

7x=25 *Divide both sides by* ‘7’

x= 25/7.

*Solving verbal (word) problems:*

*1. Read the problem. Identify the quantity asked for.*

*2. Select a letter ‘x’ or ‘y’ or ‘z’ etc., to represent the quantity asked for.*

*3. Change the word statements into symbolic statements.*

*4. write an equation in the letter selected, showing the relationship in the problem.*

*5. Solve the equation*.

*6. Check the answer to make sure that it satisfies the given condition in the problem.*

*Exercise* 2.1

*Solve the following equations*

1. x-2=7

*Given* x-2=7

*Adding* ‘2’ *on both sides.*

x-2+2=7+2

x=9

2. y+3= 10

*Given* y+3=10

*Subtract* ‘3’* on both sides.*

y+3-3=10-3

y=7

3. 6 = z+2

Given 6=z+2

*Subtract* ‘2’ *on both sides.*

6-2=z+2-2

4=z

${4}{.}{}{}\frac{3}{7}{+}{x}{=}\frac{17}{7}$

$\frac{3}{7}{+}{x}{=}\frac{17}{7}$

*Subtract* ‘3/7’ *on both sides *

$\frac{3}{7}{+}{x}{\u2013}\frac{3}{7}{=}\frac{17}{7}{\u2013}\frac{3}{7}$

${x}{=}\frac{17\u20133}{7}{=}\frac{14}{7}{=}{2}$

5. 6x=12

*Given* 6x=12

*Dividing both sides by* ‘6’

$\frac{6x}{6}=\frac{12}{6}$

x=2

${6}{.}{}{}\frac{t}{5}{=}{10}$

*Multiplying both sides by* ‘5’

$\frac{t}{5}\times 5=10\times 5$

t=50.

${7}{.}{}{}\frac{2x}{3}{=}{18}$

$Given\frac{2x}{3}=18$

*Multiplying both sides by* 3/2

$\frac{2x}{3}\times \frac{3}{2}=18\times \frac{3}{2}$

x=27.

${8}{.}{}{}{1}{.}{6}{=}\frac{y}{1.5}$

*Given* 1.6=y/1.5

*Multiplying both sides by* 1.5

$1.6\times 1.5=\frac{y}{1.5}\times 1.5$

2.4=y

9. 7x-9=16

*Given* 7x-9=16

*Add* ‘9’ *on both sides.*

7x-9+9=16+9

7x=25

*Dividing both sides by* ‘7’

$x=\frac{25}{7}$

10. 14y-8=13

*Give*n 14y-8=13

*Transposing* 8 *to R.H.S*

14y=13+8

14y=21

*Dividing both sides by* ’14’

$y=\frac{21}{14}$

$y=\frac{3}{2}$

11. 17+6p=9

*Given* 17+6p=9

*Transposing* 17 *to R.H.S*

6p=9-17

6p=-8

*Dividing both sides by* ‘6’

$\frac{6p}{6}=\u2013\frac{8}{6}$

p=-4/3

${12}{.}{}{}\frac{x}{3}{+}{1}{=}\frac{7}{15}$

${G}{i}{v}{e}{n}{}\frac{x}{3}{+}{1}{=}\frac{7}{15}$

*Transposing* ‘1’* to R.H.S*

$\frac{x}{3}{=}\frac{7}{15}{\u2013}{1}$

$\frac{x}{3}{=}\frac{7\u201315}{15}$

$\frac{x}{3}{=}{\u2013}\frac{8}{15}$

*Multiplying both sides by* 3

$\frac{x}{3}{\times}{3}{=}{\u2013}\frac{8}{15}{\times}{3}$

${x}{=}{\u2013}\frac{8}{5}$

## ncert solutions for class 8 maths chapter 2 exercise 2.2

**Exercise 2.2**

1. *If you subtract* 1/2 *from a number and multiply the result by* 1/2, *you get* 1/8. *what is the number*?

*solution: Let the required number be* ‘x’

*According to problem, *

${(}{x}{\u2013}\frac{1}{2}{)}\frac{1}{2}{=}\frac{1}{8}$

${x}{\u2013}\frac{1}{2}{=}\frac{2}{8}$

${x}{\u2013}\frac{1}{2}{=}\frac{1}{4}$

${x}{=}\frac{1}{4}{+}\frac{1}{2}$

${x}{=}\frac{1+2}{4}{=}\frac{3}{4}$

*required number is* 3/4.

2. *The perimeter of a rectangular swimming pool is* 154 m.* Its length is* 2 m *more than twice its breadth. What are the length and the breadth of the pool?*

*solution: Let breadth* = b = ‘x’ *m.*

*Then according to problem length* = l =(2x+2)* m*.

*Perimeter of pool* = 2 (l+b) =154 *m.*

2(2x+2+x)=154

2(3x+2)=154

6x+4=154

6x=154-4=150

$x=\frac{150}{6}=25$

*length of the pool* = 2x+2 = 2(25)+2=52* m.*

*breadth* = x=25 *m.*

*3. The base of an isosceles triangle is* 4/3 cm.* The perimeter of the triangle* *is* 4 ²/₁₅ *cm*. *What is the length of either of the remaining equal sides?*

*Solution : let the length of equal sides be* x *cm.*

*Given base* = 4/3.

*Perimeter* = x + x +* base* = 4 ²/₁₅ * cm.*

$2x+\frac{4}{3}=\frac{62}{15}$

*On transposing* 4/3 *to RHS we obtain.*

$2x=\frac{62}{15}\u2013\frac{4}{3}$ $2x=\frac{42}{15}$

*On dividing both sides by* 2, *we obtain*

$\frac{2x}{2}=\frac{42}{15}\times \frac{1}{2}$ $x=\frac{7}{5}=1\frac{2}{5}$

∴ *The length of equal sides is* 1²/₅ *cm.*

*4. Sum of two numbers is* 95. *If one exceeds the Other by 15, find the numbers.*

*Solution : Let one number be* ‘x’ ,

∴ *The other number will be* x+15

*According to the question,*

x+x+15=95

2x+15=95

2x=95-15

2x=80

x=80/2=40.

*The required numbers are* x=40 , x+15=55.

5.* Two numbers are in the ratio* 5:3.* If they differ by* 18. *what are the numbers?*

*Ans : *

*Let the numbers are* 5x, 3x *respectively.*

*Given difference between these numbers* =18.

5x-3x=18

2x=18

x=18/2=9

1*st number*= 5x= 5×9=45

2*nd number*= 3x=3×9=27.

6. *Three consecutive integers add up to* 51. *What are these integers?*

*Ans: Let three consecutive integers are* x,x+1* and* x+2.

*Sum of these numbers* = x + x + 1 + x + 2= 51

3x+3=51

3x=51-3=48

*On dividing both sides by* 3, *we obtain*

$\frac{3x}{3}=\frac{48}{3}$

x=16

*Hence consecutive integers are* 16, 16+1=17 *and* 16+2=18.

7 *The sum of three consecutive multiples of* 8* is* 888* find the multiples.*

*solution : Let the consecutive multiples of * 8* are* 8x, 8(x+1), 8(x+2)

*sum of these numbers*=8x+8(x+1)+8(x+2)=888.

8x+8x+8+8x+16=888.

24x+24=888

8(3x+3)=888

*On dividing both sides by* 8

$\frac{8(3x+3)}{8}=\frac{888}{8}$

3x+3=111

*On Transposing* 3 *to* *R.H.S*

3x=111-3

3x=108

*On dividing both sides by* 3.

$x=\frac{108}{3}$

x=36.

*First multiple*= 8x=8(36)=288,

*Second multiple=*8(x+1)=8(36+1)=296,

*Third multiple*=8(x+2)=8(36+2)=304.

8* Three consecutive integers are such that when they are taken in increasing order and multiplied by* 2, 3 *and* 4 *respectively, they add up to* 74. *Find these numbers.*

*solution: Let the consecutive integers are* x, x+1,x+2

*According to problem* 2x+3(x+1)+4(x+2)=74

2x+3x+3+4x+8=74

9x+11=74

*On transposing* 11 *to R.H.S*

9x=74-11

9x=63

*On dividing both sides by* 9

$\frac{9x}{9}=\frac{63}{9}$

x=7

x+1=7+1=8

x+2=7+2=9

*The numbers are* 7,8* and* 9

9 *The ages of Rahul and Haroon are in the ratio* 5:7. *Four years later the sum of their ages will b*e 56 *years what are their present ages?*

*Solution: Let the ages of rahul and haroon are* 5x,7x.

*Four years later their ages will be* 5x+4, 7x+4

*According to problem*

5x+4+7x+4=56

12x+8=56.

*Transposing* 8* to R.H.S*

12x=56-8=48

x=3.

10 *The number of boys and girls in a class are in the ratio* 7:5* the number of boys is* 8 *more than the number of girls. What is the total class strength?*

*Solution: Let the number of boys*=7x.

*Then number of girls* =5x

*According to problem *

7x=5x+8

*Transposing* 5x* to* *L.H.S*

7x-5x=8

2x=8

x=4

*Number of boys*=7x=7×4=28

*Number of girls* = 5x=5×4=20

*Total class strength*=28+20=48* students*.

11* Baichung’s father is* 26 *years younger than baichung’s grandfather and* 29 *years older than baichung. The sum of the ages of all three is* 135 *years. What is the age of each one of them?*

*Solution : Let Baichung’s father’s age is* ‘x’ *years.*

*Then Baichung’s age and baichung’s grand father’s age will be* (x-29) *years* *and* (x+26)* years respectively.*

*According to problem sum of the ages of three people is* 135 *years.*

x+x-29+x+26=135

3x-3=135

*On transposing* 3 *to R.H.S*

3x=135+3

3x=138

*On dividing both sides by* 3.

$\frac{3x}{3}=\frac{138}{3}$

x=46

*Baichung’s father’s age* =x *years*= 46 *years.*

*Baichung’s age*=(x-29)*years*=(46-29) =17 years.

*Baichung’s grand father’s age*=(x+26)=(46+26)=72* years*

12 *Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?*

*Solution: Let Ravis present age be* ‘x’ *years.*

*Fifteen years later, Ravi’s age* =4×*His present age.*

x+15=4x

*On transposing* ‘x’ *to R.H.S*

15=4x-x

15=3x

*On dividing both sides by* 3

$\frac{15}{3}=\frac{3x}{3}$

5=x

*Ravi’s present age* = 5 *years.*

13 *A rational number is such that when you multiply it by* 5/2 *and add* 2/3* to the product, you get* – 7/12 *what is the number?*

*Solution: Let the number be* x.

*According to problem,*

$\frac{5}{2}x+\frac{2}{3}=\u2013\frac{7}{12}$

*On transposing* 2/3 *to R.H.S*

$\frac{5}{2}x=\u2013\frac{7}{12}\u2013\frac{2}{3}$ $\frac{5}{2}x=\u2013\frac{15}{12}$

*on multiplying both sides by* 2/5

x=-15/12×2/5=-1/2

*The rational number is* -1/2

14 *Lakshmi is a cashier in a bank. She has currency notes of denominations Rs* 100 *Rs* 50* and Rs* 10, *respectively. The ratio of the number of these notes is* 2:3:5. * The total cash with Lakshmi is Rs* 4,00,000. *How many notes of each denomination does she have?*

*Solution: Let the common ratio between the numbers of notes of different denominations be* ‘x’.

*Numbers of Rs* 100 *notes, Rs* 50 *notes and Rs* 10* notes will be* 2x 3x *and* 5x *respectively.*

*Amount of Rs* 100 *notes=Rs*(100*2x)=200x

*Amount of Rs* 50 *notes =Rs* (50 x 3 x)=Rs150x

*Amount of Rs* 10 *notes*= *Rs* (10 x 5x )=Rs50x

*It is given the total amount is Rs* 400000.

200x+150x+50x=400000.

400x=400000.

*On dividing both sides by* 400

x=1000.

*Number of Rs* 100 *notes=*2x=2×1000=2000

*Number of Rs* 50* notes* =3x=3×1000=3000

*Number of Rs* 10 *notes*=5x=5×1000=5000

15* I have a total of Rs* 300* in coins of denomination Re* 1,* Rs* 2* and Rs* 5.* The number of Rs* 2 *coins is* 3* times the number of Rs* 5 *coins. The total number of coins is* 160.* How many coins of each denomination are with me?*

*Solution: Let the number of Rs* 5 *coins be* X.

*Number of Rs* 2 *coins=* 3 ×*number of Rs* 5 *coins* 3x

*Number of Re* 1 *coins* =160 – *(Number of coins of Rs* 5* and of Rs* 2)

=160-(3x+x)=160-4x

*Amount of Re* 1 *coins* = *Rs*[1*(160-4x)]=Rs(160-4x)

*Amount of Rs* 2* coins*=*Rs *(2×3x)=*Rs* 6x

*Amount of Rs* 5* coins=Rs*(5x)=*Rs* 5x

*It is given that the total amount is Rs* 300.

160-4x+6x+5x=300

160+7x=300

*On transposing* 160 *to R.H.S,*

7x=300-160

7x=140

*on dividing both sides by* 7

x=20

*Number of Re*1* coins*=160-4x=160-4×20=160-80=80

*Number of R*s 2 *coin*s=3x=3×20=60

*Number of R*s 5* coins*=x=20.

16* The organisers of an essay competition decide the a winner in the competition gets a prize of Rs* 100 *and a participant who does not win get a price of Rs* 25. *The total prize money distributed is Rs* 3,000. *Find the number of winners, if the total number of participants is* 63.

*Solution: Let the number of winner be* X.

*Therefore, the number of participants who did not pin will be* 63 – X.

*Amount given to the winner*=Rs 100x

*Amount given to the participant who did not win* =Rs[25(63-x)]=Rs(1575-25x)

*According to the given problem*

*On transposing* 1575 *to R.H.S*

75x=3000-1575

75x=1425

*on dividing both sides by* 75

$\frac{75x}{75}=\frac{1425}{75}$

x=19

*Number of winners=*19

**Exercise 2.3**

*Solve the following equations and check your results.*

1 3x=2x+18

Answer: Given 3x=2x+18

on transposing 2x to L.H.S

3x-2x=18

x=18

L.H.S=3x=3(18)=54

R.H.S=2x+18=2(18)+18=36+18=54

L.H.S=R.H.S

The result obtained above is correct.

2 5t-3=3t-5

Answer: Transposing 3t to L.H.S and -3 to R.H.S

5t-3t=-5+3

2t=-2

Dividing on both sides by 2

2t/2=-2/2

t=-1

L.H.S=5t-3=5(-1)-3=-5-3=-8

R.H.S=3t-5=3(-1)-5=-3-5=-8

L.H.S=R.H.S

The result obtained above is correct

3 5x+9=5+3x

Answer: Given 5x+9=5+3x

Transposing 3x to L.H.S and 9 to R.H.S

5x-3x=5-9

2x=-4

Dividing on both sides by 2

$\frac{2x}{2}=\frac{\u20134}{2}$

x=-2

4 4z+3=6+2z

Answer: Given 4z+3=6+2z

Transposing 2z to L.H.S and 3 to R.H.S

4z-2z=6-3

2z=3

On dividing both sides by 2

$z=\frac{3}{2}$

L.H.S=4z+3=4(3/2)+3=2(3)+3=6+3=9

R.H.S=6+2z=6+2(3/2)=6+3=9

L.H.S=R.H.S

The result obtained above is correct.

5 2x-1=14-x

Answer: Given 2x-1=14-x

Transposing -x to L.H.S and -1 to R.H.S

2x+x=14+1

3x=15

Dividing on both sides by 3

$\frac{3x}{3}=\frac{15}{3}$

x=5

L.H.S=2x-1=2(5)-1=10-1=9

R.H.S=14-x=14-5=9

L.H.S=R.H.S

The result obtained above is correct.

6 8x+4=3(x-1)+7

Answer: Given 8x+4=3(x-1)+7

8x+4=3x-3+7

8x+4=3x+4

On transposing 3x to L.H.S and 4 to R.H.S

8x-3x=4-4

4x=0

x=0

L.H.S=8x+4=8(0)+4=4

R.H.S=3(x-1)+7=3(0-1)+7=-3+7=4

L.H.S=R.H.S

The result obtained above is correct.

${7}{.}{}{x}{=}\frac{4}{5}{(}{x}{+}{10}{)}$

Answer: Given

$x=\frac{4}{5}(x+10)$

On multiplying both sides by 5

5x=4(x+10)

5x= 4x+40

Transposing 4x to L.H.S

5x-4x=40

x=40

L.H.S=x=40

$R.H.S=\frac{4}{5}(x+10)=\frac{4}{5}(40+10)=\frac{4}{5}\times 50=4\times 10=40$

L.H.S=R.H.S

The result obtained above is correct.

${8}{.}{}\frac{2x}{3}{+}{1}{=}\frac{7x}{15}{+}{3}$

Answer: Given

$\frac{2x}{3}+1=\frac{7x}{15}+3$

On transposing 7x/15 to L.H.S and 1 to R.H.S.

$\frac{2x}{3}\u2013\frac{7x}{15}=3\u20131$ $\frac{10x\u20137x}{15}=2$

$\frac{3x}{15}=2\phantom{\rule{0ex}{0ex}}\frac{x}{5}=2\phantom{\rule{0ex}{0ex}}$

Multiplying on both sides with 5

x=10

$L.H.S.=\frac{2x}{3}+1=\frac{2\left(10\right)}{3}+1=\frac{20}{3}+1=\frac{23}{3}$ $R.H.S=\frac{7x}{15}+3=\frac{7\left(10\right)}{15}+3=\frac{14}{3}+3=\frac{14+9}{3}=\frac{23}{3}$

L.H.S=R.H.S

The result obtained above is correct.

${9}{}{}{2}{y}{+}\frac{5}{3}{=}\frac{26}{3}{\u2013}{y}$

Transposing y to L.H.S and 5/3 toR.H.S

${2}{y}{+}{y}{=}\frac{26}{3}{\u2013}\frac{5}{3}$

${3}{y}{=}\frac{26\u20135}{3}$

${3}{y}{=}\frac{21}{3}$

3y=7

dividing both sides by 3

${y}{=}\frac{7}{3}$

${L}{.}{H}{.}{S}{.}{=}{2}{y}{+}\frac{5}{3}{=}{2}{\left(}\frac{7}{3}{\right)}{+}\frac{5}{3}{=}\frac{19}{3}$

${R}{.}{H}{.}{S}{.}{=}\frac{26}{3}{\u2013}{y}{=}\frac{26}{3}{\u2013}\frac{7}{3}{=}\frac{26\u20137}{3}{=}\frac{19}{3}$

L.H.S.=R.H.S.

Hence, the result obtained above is correct.

${10}{}{}{3}{m}{=}{5}{m}{\u2013}\frac{8}{5}$

${G}{i}{v}{e}{n}{}{3}{m}{=}{5}{m}{\u2013}\frac{8}{5}$

Transposing 5m to L.H.S,

${3}{m}{\u2013}{5}{m}{=}{\u2013}\frac{8}{5}$

${\u2013}{2}{m}{=}{\u2013}\frac{8}{5}$

Dividing both sides by -2,

${m}{=}\frac{4}{5}$

${L}{.}{H}{.}{S}{=}{3}{m}{=}{3}{\left(}\frac{4}{5}{\right)}{=}\frac{12}{5}$

${R}{.}{H}{.}{S}{=}{5}{m}{\u2013}\frac{8}{5}{=}{5}{\left(}\frac{4}{5}{\right)}{\u2013}\frac{8}{5}{=}\frac{20\u20138}{5}{=}\frac{12}{5}$

L.H.S.=R.H.S.

Hence, the result obtained above is correct.

**Exercise:2.4**

1 Amina thinks of a number and subtract 5|2 from it. She multiplies results by 8. The result now obtained if 3 times the same number she thought of. What is the number?

Solution: Let the number be x.

According to the given problem,

${8}{(}{x}{\u2013}\frac{5}{2}{)}{=}{3}{x}$

8x-20=3x

Transposing 3x to L.H.S. and -22 to R.H.S

8x-3x=20

5x=20

Dividing both sides by 5.

x=4

Hence, the number is 4

2 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution: Let the numbers be x and 5x. According to the problem.

5x+21=2(x+21)

⇒5x+21=2x+42

Transposing to 2x to L.H.S. and 21 to R.H.S

5x-2x=42-21

3x=21

Dividing both sides by 3

x=7

Hence the number are x=7 and 5x=35.

3 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. what is the two digit number?

Solution: Let x be tens place digit. Then units place digit =9-x.

Then original number=10x+(9-x)1=9x+9

After interchanging the digits

New number=10(9-x)+(x)1=90-9x

According to the problem,

New number=original number+27

90-9x=9x+9+27

90-9x=36+9x

Transposing 9X to R.H.S. and 36 to L.H.S.

90-36=9x+9x

54=18x

Dividing both sides by 18,

3=x and 9-x=9-3=6

Tens place digit =6, units place digit =3

The two digit number =9x+9=9(3)+9=36.

4 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Solution: Let the digits at tens place and ones place be x and 3x respectively.

Then original number = 10x + 3x=13x

On interchanging the digits, the digits at ones place and tens place will be x and 3x respectively.

Number after interchanging =10×3x+x=30x+x=31x

According to the given question,

Original number+new number=l 88

13 x + 31x=88

44x=88

Dividing both sides by 44,

x=2

our original number=13x=13×2=26

By considering the tens place and ones place as 3x and x respectively,

The two digit number is obtained is 62.

Therefore the two digit number may be 26 or 62.

5 shobo’s mothers present age is 6 times shobo’s present age. Shobo’s age 5 years from now will be one third of his mother’s present age what are their present ages?

Solution: Let shobo’s age be x years.

Then his mother’s age will be 6x years.

According to the problem,

${A}{f}{t}{e}{r}{}{5}{y}{r}{s}{}{s}{h}{o}{b}{o}{\u2018}{s}{}{a}{g}{e}{=}\frac{shobo\u2018smotherpresentage}{3}$

${x}{+}{5}{=}\frac{6x}{3}$

x+5=2x

Transposing x to R.H.S,

5=2x-x

5=x

6x=6(5)=30.

∴ The present ages of shobo and shobo’s mother will be 5 years and 30 years respectively.

6 There is a narrow rectangular plot, reserved for school, in Mohuli village. The length and breadth of the plot are in the ratio 11:4. At the rate of Rs 100 per metre it will cost the village panchayat at Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution:

Given length bredth ratio is 11:4

Let length=11x, breadth=4x.

Perimeter of the plot =2(11x+4x)=2(15x)=30x m

Given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75000

∴ 100×perimeter=75000

100×30x=75000

3000x=75000

x=25

Length of the plot=11x=11(25)=275 m.

Breadth=4x=4(25)=100 m.

7 Hasan buys two kinds of cloth material for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that cost him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12 % and 10% profit respectively. His total sale is Rs 36,660. How much trouser material did you buy?

Solution: Given for every 2m of trouser material he buys 3m of shirt material.

Let 2x m trouser material and 3x m shirt material be bought by him.

Per metre selling price of trouser material=

${(}{90}{+}{90}{\times}\frac{12}{100}{)}{=}{R}{s}{.}{}{100}{.}{80}$

Per metre selling price of shirt material=

${(}{50}{+}{50}{\times}\frac{10}{100}{)}{=}{R}{s}{.}{}{55}$

Given that total amount of selling=36,660.

(100.8)×(2x)+55×(3x)=36660.

201.60x+165x=36660

366.60x=36660

Dividing both sides by 366.60

x=100

Trouser material=2x =2(100)=200m.

8 Half of a herd of deer are grazing in the field and three fourth of remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Solution: Let the number of deer =x.

Number of deer grazing in the field=x/2

Number of deer playing nearby=3/4×Number of remaining deer.

${=}\frac{3}{4}{\times}{[}{x}{\u2013}{(}\frac{x}{2}{\left)}{\right]}{=}\frac{3}{4}{\times}\frac{x}{2}{=}\frac{3x}{8}$

Number of deer drinking water from the pond =9

${x}{\u2013}{(}\frac{x}{2}{+}\frac{3x}{8}{)}{=}{9}$

${x}{\u2013}\frac{7x}{8}{=}{9}$

$\frac{x}{8}{=}{9}$

Multiplying both sides by 8,

x=72.

The total number of deer in the herd is 72.

9 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Solution: Let The grand daughters age be x years then grandfather’s age will be 10x years.

According to the problem,

Grandfather’s age = Granddaughter’s age +54 years

10x=x+54

Transposing x to L.H.S,

10x-x=54

9x=54

x=6.

Granddaughter’s age =x years=6 years.

Grandfather’s age=10x=(10×6) years=60years.

10 Aman’s age is 3 times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Solution: Let aman’s son age =x years.

Then aman’s age=3x years.

Ten years ago their ages was x-10, 3x-10 years respectively.

According to problem 10 years ago

Aman’s age=5×aman’s son’s age.

3x-10=5(x-10)

3x-10=5x-50

Transposing 3x to R.H.S and 50 to L.H.S,

50-10=5x-3x

40=2x

x=20

Aman’s son’s age=x years=20 years.

Aman’s age =3x years=(3×20) years=60 years.

**Exercise 2.5**

Solve the following linear equations.

${1}{}{}{}\frac{x}{2}{\u2013}\frac{1}{5}{=}\frac{x}{3}{+}\frac{1}{4}$

${G}{i}{v}{e}{n}{}\frac{x}{2}{\u2013}\frac{1}{5}{=}\frac{x}{3}{+}\frac{1}{4}$

Solution: LCM of denominators 2,3,4 and 5 is 60

Multiply both sides by 60.

${60}{(}\frac{x}{2}{\u2013}\frac{1}{5}{)}{=}{60}{(}\frac{x}{3}{+}\frac{1}{4}{)}$

⇒30x-12=20x+15

Transposing 20x to L.H.S and 12 to R.H.S

⇒30x-20x=15+12

10x=27

${x}{=}\frac{27}{10}$

${2}{}{}{}\frac{n}{2}{\u2013}\frac{3n}{4}{+}\frac{5n}{6}{=}{21}$

Solution: LCM of denominators 2,4 and 6 is 12

Multiply both sides by 12

${12}{(}\frac{n}{2}{\u2013}\frac{3n}{4}{+}\frac{5n}{6}{)}{=}{12}{\times}{21}$

6n-9n+10n=252

7n=252

${n}{=}\frac{252}{7}$

n=36

${3}{}{}{x}{+}{7}{\u2013}{\left(}\frac{8x}{3}{\right)}{=}\frac{17}{6}{\u2013}\frac{5x}{2}$

${G}{i}{v}{e}{n}{}{}{x}{+}{7}{\u2013}{\left(}\frac{8x}{3}{\right)}{=}\frac{17}{6}{\u2013}\frac{5x}{2}$

LCM of denominators 2,3and 6 is 6

Multiply both sides by 6

${6}{[}{x}{+}{7}{\u2013}{(}\frac{8x}{3}{\left)}{\right]}{=}{6}{[}\frac{17}{6}{\u2013}\frac{5x}{2}{]}$

6x+42-16x=17-15x

-10x+42=17-15x

Transpose 15x to L.H.S and 42 to R.H.S

-10x+15x=17-42

5x=-25

x=-5

${4}{}{}\frac{(x\u20135)}{3}{=}\frac{(x\u20133)}{5}$

${G}{i}{v}{e}{n}{}\frac{(x\u20135)}{3}{=}\frac{(x\u20133)}{5}$

LCM of denominators 3 and 5 is 15

Multiply both sides by 15

${15}{\times}\frac{(x\u20135)}{3}{=}\frac{(x\u20133)}{5}{\times}{15}$

5(x-5)=3(x-3)

5x-25=3x-9

Transposing 3x to L.H.S and 25 to R.H.S

5x-3x=25-9

2x=16

x=8

${5}{}{}\frac{(3t\u20132)}{4}{\u2013}\frac{(2t+3)}{3}{=}\frac{2}{3}{\u2013}{t}$

S0lution:

${G}{i}{v}{e}{n}\frac{(3t\u20132)}{4}{\u2013}\frac{(2t+3)}{3}{=}\frac{2}{3}{\u2013}{t}$

LCM of denominators 4 and 3 is 12

Multiply both sides by 12

${12}{[}\frac{(3t\u20132)}{4}{\u2013}\frac{(2t+3)}{3}{]}{=}{12}{[}\frac{2}{3}{\u2013}{t}{]}$

9t-6-8t-12=8-12t

t-18=8-12t

12t+t=8+18

13t=26

t=2

${6}{}{}{}{m}{\u2013}\frac{(m\u20131)}{2}{=}{1}{\u2013}\frac{(m\u20132)}{3}$

Solution:Given

${m}{\u2013}\frac{(m\u20131)}{2}{=}{1}{\u2013}\frac{(m\u20132)}{3}$

LCM of denominators 2 and 3 is 6

Multiply both sides by 6

${6}{[}{m}{\u2013}\frac{(m\u20131)}{2}{]}{=}{6}{[}{1}{\u2013}\frac{(m\u20132)}{3}{]}$

6m-3m+3=6-2m+4

3m+3=10-2m

3m+2m=10-3

5m=7

${m}{=}\frac{7}{5}$

Simplify and solve the following linear equations.

7 3(t-3)=5(2t+1)

Solution: 3t-9=10t+5

3t-10t=5+9

-7t=14

t=-2

8 15(y-4)-2(y-9)+5(y+6)=0

Solution: 15y-60-2y+18+5y+30=0

18y-12=0

18y=12

${y}{=}\frac{12}{18}$

${y}{=}\frac{2}{3}$

9 3(5z-7)-2(9z-11)=4(8z-13)-17

Solution: 15z-21-18z+22=32z-52-17

-3z+1=32z-69

69+1=32z+3z

70=35z

Dividing both sides by 35

2 =z

10 0.25(4f-3)=0.05(10f-9)

Solution : 1f-0.75=0.5f-0.45

f-0.5f=0.75-0.45

0.5f=0.3

Dividing both sides by 0.5

f=3 /5=0.6

**Exercise 2.6**

Solve the following equations.

${1}{}{}\frac{(8x\u20133)}{3x}{=}{2}$

Solution:

${G}{i}{v}{e}{n}{}\frac{(8x\u20133)}{3x}{=}{2}$

on multiplying both sides by 3x.

${3}{x}{\left[}\frac{8x\u20133}{3x}{\right]}{=}{3}{x}{\left(}{2}{\right)}$

8x-3=6x

8x-6x=3

2x=3

${x}{=}\frac{3}{2}$

${2}{}{}\frac{9x}{7\u20136x}{=}{15}$

Solution :

${G}{i}{v}{e}{n}{}\frac{9x}{7\u20136x}{=}{15}$

on multiplying both sides by 7-6x

${(}{7}{\u2013}{6}{x}{)}{\left[}\frac{9x}{7\u20136x}{\right]}{=}{(}{7}{\u2013}{6}{x}{)}{15}$

9x=105-90x

9x+90x=105

99x=105

${x}{=}\frac{105}{99}{=}\frac{35}{33}$

${3}{}{}\frac{z}{(z+15)}{=}\frac{4}{9}$

Solution :

${G}{i}{v}{e}{n}{}\frac{z}{(z+15)}{=}\frac{4}{9}$

On multiplying both sides by 9(z+15)

9z=4(z+15)

9z=4z+60

9z-4z=60

5z=60

On dividing both sides by 5

z=12

${4}{}{}\frac{(3y+4)}{(2\u20136y)}{=}{\u2013}\frac{2}{5}$

Solution:

${G}{i}{v}{e}{n}{}\frac{(3y+4)}{(2\u20136y)}{=}{\u2013}\frac{2}{5}$

On multiplying both sides by 5(2-6y)

5(3y+4)=-2(2-6y)

15y+20=-4+12y

15y-12y=-4-20

3y=-24

On dividing both sides by 3

y=-8

${5}{}{}\frac{(7y+4)}{(y+2)}{=}{\u2013}\frac{4}{3}$

Solution:

${G}{i}{v}{e}{n}{}\frac{(7y+4)}{(y+2)}{=}{\u2013}\frac{4}{3}$

On multiplying both sides by 3(y+2)

3(7y+4)=-4(y+2)

21y+12=-4y-8

21y+4y=-8-12

25y=-20

On dividing both sides by 25

${y}{=}{\u2013}\frac{20}{25}{=}{\u2013}\frac{4}{5}$

6 The ages of Hari and Harry are in the ratio 5: 7. Four years from now the ratio of their ages will be 3:4 find their present ages.

Solution:

Let the common ratio between the ages be x.

Then heri’s age and herry’s age will be 5x years and 7 years respectively.

And four years later, their ages will be (5x+4) and (7x + 4) respectively.

According to the problem,

$\frac{5x+4}{7x+4}{=}\frac{3}{4}$

.

4(5x+4)=3(7x+4)

20x+16=21x+12

20x-21x=12-16

-x=-4

x=4

Hari’s age =5x =5(4)=20 years

Harry’s age =7x =7(4)=28 years

∴Hari’s age and Harry’s age are 20 years and 28 years respectively.

7 The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3 /2 find the rational number.

Solution : Let the numerator of the rational number be x.

Then its denominator will be x+ 8 the rational number will be x /x+ 8

According to the problem,

$\frac{x+17}{x+8\u20131}{=}\frac{3}{2}$

$\frac{x+17}{x+7}{=}\frac{3}{2}$

2(x+17)=3(x+7)

2x+34=3x+21

2x-3x=21-34

-x=-13

x=13.

Numerator of rational number=x=13

Denominator of rational number=x+8=13+8=21

Required rational number=13 /21.